AP DSC School Assistant Math Model Test 3

Concept:Finding derivatives and critical points of a trigonometric function to determine maxima and minima within a closed interval.Explanation:Given function: $f(x) = \sin x + \frac{1}{2}\cos 2x$ on $[0, \frac{\pi}{2}]$.1. Differentiate: $f'(x) = \cos x + \frac{1}{2}(-2\sin 2x) = \cos x - \sin 2x$. Hence option (A) is correct.2. Critical points occur where $f'(x)=0$: $\cos x - \sin 2x = 0 \Rightarrow \cos x = \sin 2x = 2\sin x\cos x$. If $\cos x \neq 0$, divide: $1 = 2\sin x \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}$. If $\cos x = 0$, then $x = \frac{\pi}{2}$ (since interval includes $\frac{\pi}{2}$). Both $x=\frac{\pi}{6}$ and $x=\frac{\pi}{2}$ lie in $[0,\frac{\pi}{2}]$, so option (B) is correct.3. Evaluate $f(x)$ at endpoints and critical points: $f(0) = 0 + \frac{1}{2}\cdot 1 = 0.5$ $f(\frac{\pi}{6}) = \frac{1}{2} + \frac{1}{2}\cos\frac{\pi}{3} = \frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4}$ $f(\frac{\pi}{2}) = 1 + \frac{1}{2}\cos\pi = 1 - \frac{1}{2} = 0.5$ Minimum value = $0.5$, maximum value = $\frac{3}{4}$. So option (D) is correct, but option (C) is false (minimum is not 2).Thus statements (A), (B), (D) are true.Answer:Option A: (A), (B) and (D) only.